Answer:
(a) maximum positive reaction at A = 64.0 k
(b) maximum positive shear at A = 32.0 k
(c) maximum negative moment at C = -540 k·ft
Explanation:
Given;
dead load Gk = 400 lb/ft
live load Qk = 2 k/ft
concentrated live load Pk =8 k
(a) from the influence line for vertical reaction at A, the maximum positive reaction is
= 2*(8) +(1/2(20 - 0)* (2))*(2 + 0.4) = 64 k
See attachment for the calculations of (b) & (c) including the influence line
Answer with Explanation:
The general equation of simple harmonic motion is
where,
A is the amplitude of motion
is the angular frequency of the motion
is known as initial phase
part 1)
Now by definition of velocity we have
part 2)
Now by definition of acceleration we have
part 3)
The angular frequency is related to Time period 'T' as
where
is the angular frequency of the motion of the particle.
Part 4) The acceleration and velocities are plotted below
since the maximum value that the sin(x) and cos(x) can achieve in their respective domains equals 1 thus the maximum value of acceleration and velocity is 
and 
respectively.
Less, if it’s too big: hard to control and maneuverability for shooting wouldn’t be that good. a smaller wheelchair allows for faster movement and control, along with easier shooting and upper body movement
Answer:
a) V =10¹¹*(1.5q₁ + 3q₂)
b) U = 1.34*10¹¹q₁q₂
Explanation:
Given
x₁ = 6 cm
y₁ = 0 cm
x₂ = 0 cm
y₂ = 3 cm
q₁ = unknown value in Coulomb
q₂ = unknown value in Coulomb
A) V₁ = Kq₁/r₁
where r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m
V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁
V₂ = Kq₂/r₂
where r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m
V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂
The electric potential due to the two charges at the origin is
V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)
B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows
U = Kq₁q₂/r₁₂
where
r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m
then
U = 9*10⁹q₁q₂/(3√5/100)
⇒ U = 1.34*10¹¹q₁q₂
